Question

If ${}^n{C}_9{=}^n{C}_{8,}$ then find ${}^n{C}_{17}$.

Answer 1

${}^n{C}_9{=}^n{C}_8$ ${\because }^n{C}_r=\frac{n!}{r!(n-r)!}$

$\frac{n!}{9(n-9)!}=\frac{n!}{8!(n-8)!}$

$\Rightarrow \frac{n!(n-8)!}{n!(n-9)!}=\frac{9!}{8!}$

$\Rightarrow \frac{(n-8)!}{(n-9)!}=\frac{9!}{8!}$

$\Rightarrow \frac{(n-8)(n-8-1)!}{(n-9)!}=\frac{9\times 8!}{8!}$

$\Rightarrow \frac{(n-8)(n-9)!}{(n-9)!}=9$

$\Rightarrow n-8=9$

$\Rightarrow n=17$

${}^n{C}_{17}$

putting n=17

${=}^{17}{C}_{17}$

$=\frac{17!}{17!(17-17)!}$

$=\frac{17!}{17!\times 0!}$

$=\frac{17!}{17!\times 1}$

$=\frac{1}{1}$

${\therefore }^n{C}_{17}=1$